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2016 AMC 10A Problems/Problem 1: Difference between revisions

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==Solution 1==
==Solution 1==


Factoring out <math>10!</math> from the numerator and cancelling out <math>9!</math> from the numerator and the denominator, we have <cmath>\frac{11!-10!}{9!} = \frac{11 \cdot 10! - 1 \cdot 10!}{9!} =  \frac{(10!) \cdot (11 - 1)}{9!} = 10 \cdot 10 =\boxed{\textbf{(B)}\;100}.</cmath>
We can use subtraction of fractions to get <cmath>\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{\textbf{(B)}\;100}.</cmath>


==Solution 2==
==Solution 2==


We can use subtraction of fractions to get <cmath>\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{\textbf{(B)}\;100}.</cmath>
Factoring out <math>9!</math> gives <math>\frac{11!-10!}{9!} = \frac{9!(11 \cdot 10 - 10)}{9!} = 110-10=\boxed{\textbf{(B)}~100}</math>.




==Solution 3==
==Solution 3==
<math>\dfrac{11!-10!}{9!}</math>
consider 10 as n
<math>\dfrac{(n+1)!-n!}{(n-1)!}</math>
simpify
<math>\dfrac{(n+1)n!+(-1)n!}{(n-1)!}</math> = <math>\dfrac{n(n!)}{(n-1)!}</math> = <math>\dfrac{n(n(n-1)!)}{(n-1)!}</math> = <math>\dfrac{n(n)(1)}{1}</math> = <math>\dfrac{n^2}{1}</math>
subsitute n as 10 again 
<math>\dfrac{10^2}{1}</math>
answer is <math>10^2</math> which is <math>\boxed{\textbf{(B)}~100}</math>.
==Solution 4==
We are given the equation <math>\frac{11!-10!}{9!}</math>
This is equivalent to <math>\frac{11(10!) - 1(10!)}{9!}</math>
Simplifying, we get <math>\frac{(11-1)(10!)}{9!}</math>, which equals <math>10 \cdot 10</math>.
Therefore, the answer is <math>10^2</math> = <math>\boxed{\textbf{(B)}~100}</math>.
~TheGoldenRetriever
==Solution 5 (This is a joke)==
Let
<cmath>
I_n := \int_0^\infty x^n e^{-x} \, dx = n!
</cmath>
(the Gamma–integral).
Then
<cmath>
\frac{11! - 10!}{9!} = \frac{I_{11} - I_{10}}{I_9} = \frac{\int_0^\infty e^{-x} (x^{11} - x^{10}) \, dx}{I_9}.
</cmath>
Integration by parts on the numerator with \( u = x^{11} - x^{10} \), \( dv = e^{-x} \, dx \) (so \( du = (11x^{10} - 10x^9) \, dx \), \( v = -e^{-x} \)) gives
<cmath>
\int_0^\infty e^{-x} (x^{11} - x^{10}) \, dx = \left[ -e^{-x} (x^{11} - x^{10}) \right]_0^\infty + \int_0^\infty e^{-x} (11x^{10} - 10x^9) \, dx = 11 I_{10} - 10 I_9,
</cmath>
since the boundary term vanishes.
Hence
<cmath>
\frac{I_{11} - I_{10}}{I_9} = \frac{11 I_{10}}{I_9} - 10.
</cmath>
Do another integration by parts to relate \( I_{10} \) and \( I_9 \):
<cmath>
I_{10} = \int_0^\infty x^{10} e^{-x} \, dx = \left[ -x^{10} e^{-x} \right]_0^\infty + 10 \int_0^\infty x^9 e^{-x} \, dx = 10 I_9.
</cmath>
Therefore
<cmath>
\frac{I_{11} - I_{10}}{I_9} = 11 \cdot 10 - 10 = 100.
</cmath>
<math>\boxed{(B) 100}</math>.
~Pinotation
~Minorly Edited by OffBrandCab
==Video Solution (HOW TO THINK CREATIVELY!!!)==
https://youtu.be/r5G98oPPyNM
~Education, the Study of Everything
==Video Solution==
https://youtu.be/VIt6LnkV4_w
https://youtu.be/CrS7oHDrvP8
~savannahsolver
==Video Solution (FASTEST METHOD!)==
https://youtu.be/jowREGsZaTs
~Veer Mahajan


Factoring out <math>9!</math> gives <math>\frac{11!-10!}{9!} = \frac{9!(11 \cdot 10 - 10)}{9!} = 110-10=\boxed{\textbf{(B)}~100}</math>.


==See Also==
==See Also==
{{AMC10 box|year=2016|ab=A|before=|num-a=2}}
{{AMC10 box|year=2016|ab=A|before=First Problem|num-a=2}}
{{AMC12 box|year=2016|ab=A|before=|num-a=2}}
{{AMC12 box|year=2016|ab=A|before=First Problem|num-a=2}}
{{MAA Notice}}
{{MAA Notice}}
[[Category: Introductory Algebra Problems]]

Latest revision as of 22:22, 26 October 2025

Problem

What is the value of $\dfrac{11!-10!}{9!}$?

$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$

Solution 1

We can use subtraction of fractions to get \[\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{\textbf{(B)}\;100}.\]

Solution 2

Factoring out $9!$ gives $\frac{11!-10!}{9!} = \frac{9!(11 \cdot 10 - 10)}{9!} = 110-10=\boxed{\textbf{(B)}~100}$.


Solution 3

$\dfrac{11!-10!}{9!}$ consider 10 as n $\dfrac{(n+1)!-n!}{(n-1)!}$ simpify $\dfrac{(n+1)n!+(-1)n!}{(n-1)!}$ = $\dfrac{n(n!)}{(n-1)!}$ = $\dfrac{n(n(n-1)!)}{(n-1)!}$ = $\dfrac{n(n)(1)}{1}$ = $\dfrac{n^2}{1}$ subsitute n as 10 again $\dfrac{10^2}{1}$

answer is $10^2$ which is $\boxed{\textbf{(B)}~100}$.

Solution 4

We are given the equation $\frac{11!-10!}{9!}$

This is equivalent to $\frac{11(10!) - 1(10!)}{9!}$ Simplifying, we get $\frac{(11-1)(10!)}{9!}$, which equals $10 \cdot 10$.

Therefore, the answer is $10^2$ = $\boxed{\textbf{(B)}~100}$.

~TheGoldenRetriever

Solution 5 (This is a joke)

Let \[I_n := \int_0^\infty x^n e^{-x} \, dx = n!\] (the Gamma–integral).

Then \[\frac{11! - 10!}{9!} = \frac{I_{11} - I_{10}}{I_9} = \frac{\int_0^\infty e^{-x} (x^{11} - x^{10}) \, dx}{I_9}.\] Integration by parts on the numerator with \( u = x^{11} - x^{10} \), \( dv = e^{-x} \, dx \) (so \( du = (11x^{10} - 10x^9) \, dx \), \( v = -e^{-x} \)) gives \[\int_0^\infty e^{-x} (x^{11} - x^{10}) \, dx = \left[ -e^{-x} (x^{11} - x^{10}) \right]_0^\infty + \int_0^\infty e^{-x} (11x^{10} - 10x^9) \, dx = 11 I_{10} - 10 I_9,\] since the boundary term vanishes.

Hence \[\frac{I_{11} - I_{10}}{I_9} = \frac{11 I_{10}}{I_9} - 10.\] Do another integration by parts to relate \( I_{10} \) and \( I_9 \): \[I_{10} = \int_0^\infty x^{10} e^{-x} \, dx = \left[ -x^{10} e^{-x} \right]_0^\infty + 10 \int_0^\infty x^9 e^{-x} \, dx = 10 I_9.\] Therefore \[\frac{I_{11} - I_{10}}{I_9} = 11 \cdot 10 - 10 = 100.\]

$\boxed{(B) 100}$.

~Pinotation

~Minorly Edited by OffBrandCab

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/r5G98oPPyNM

~Education, the Study of Everything


Video Solution

https://youtu.be/VIt6LnkV4_w


https://youtu.be/CrS7oHDrvP8

~savannahsolver

Video Solution (FASTEST METHOD!)

https://youtu.be/jowREGsZaTs

~Veer Mahajan


See Also

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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